Time Based Key-Value Store - Leetcode 981 Java Solution

Clean Code Binary Search Solution in Java with detailed explanation and complexity analysis.

Avatar Posted by Omar YAYA on March 10, 2024 · 7 mins read


Problem Description

LeetCode-981. Time Based Key-Value Store.

Design a time-based key-value data structure that can store multiple values for the same key at different time stamps and retrieve the key’s value at a certain timestamp.

Implement the TimeMap class:

  • TimeMap() Initializes the object of the data structure.
  • void set(String key, String value, int timestamp) Stores the key key with the value value at the given time timestamp.
  • String get(String key, int timestamp) Returns a value such that set was called previously, with timestamp_prev <= timestamp. If there are multiple such values, it returns the value associated with the largest timestamp_prev. If there are no values, it returns “”.


In the Java world, the way I would think about solving this problem is a 3-Dimensional Map. We not only need to keep a key-value pair, we also need to track it versus time.

The key does not change with time, but the value does. So we need to have a list of values for a given key at different timestamps.

Real-world example

This could be a very good application for a weather app. The key would be the city name, and the values would be the temperature on a given day.


set("berlin", "15", 1);
set("paris", "17", 1);
set("london", "9", 2);
set("paris", "19", 2);
get("berlin", 1); // get weather for Berlin @ 1 = {"15"}
set("paris", "12", 6);
get("paris", 3); // no value for Paris @ 3, closest is @ 2 ={"19"}
set("berlin", "17", 2);
set("london", "10", 4);
get("london", 10); // no value for London @ 10, closest is @ 4 ={"10"}
set("london", "3", 5);

According to the problem description, get() should return “a value such that set was called previously, with timestamp_prev <= timestamp”, which is why we returned the value @ 2 for Paris and @ 4 for London.



We will create a Map whose key is city name and its value is a List containing the pair of value & timestamp. With each get() call, we loop through all the values for the given key, until we get the one at the closest timestamp.

Map<String, List<Pair<Integer, String>>> lookup;

With every get() call, we will go through all the timestamps, so the time complexity will be O(n) where n is the number of values for a given key.


However, if we sort those values based on their timestamps, we could easily perform binary search, so the complexity becomes O(log(n)).

The preferred data structure in Java for binary search is the TreeMap. So our lookup will become:

Map<String, TreeMap<Integer, String>> lookup;

What happens when set() is called?

  • Get the sorted list of values for a given key with their corresponding timestamps
  • Add a new entry to the sorted list
  • Update the lookup table with the new data

What happens when get() is called?

  • Get the sorted list of values for a given key if it exists in the lookup table
  • Perform a binary search to find the closest timestamp
  • Return the value at the closest timestamp
  • If no value is found, return an empty string


  • Time complexity: O(1) to get the key from the map (in our example, city), then O(log(n)) to get the value at a given timestamp.

So overall time complexity is O(log(n)).

  • Space complexity: We are not using any “EXTRA” space. The data is stored in the same object that we query, so the space complexity is O(1).


class TimeMap {

    Map<String, TreeMap<Integer, String>> lookup;
    public TimeMap() {
        lookup = new HashMap<>();
    public void set(String key, String value, int timestamp) {
        TreeMap<Integer, String> sortedValuesForKey = lookup.getOrDefault(key, new TreeMap<>());
        sortedValuesForKey.put(timestamp, value);
        lookup.put(key, sortedValuesForKey);

    public String get(String key, int timestamp) {
        if(!lookup.containsKey(key)) return "";
        TreeMap<Integer, String> sortedValuesForKey = lookup.get(key);
        Integer closestTimestamp = sortedValuesForKey.floorKey(timestamp);
        if(closestTimestamp == null) return "";

        return sortedValuesForKey.getOrDefault(closestTimestamp, "");

sortedValuesForKey.floorKey(timestamp); returns the greatest key less than or equal to the given key, or null if there is no such key; this is the main binary search operation in the code, and it is exactly what we need to find the closest timestamp.

I hope you enjoyed this post.

If you have any questions, please feel free to reach out via LinkedIn Linkedin.com/omaryaya